Q 124 - Intermediate Algebra

Question

Write an application problem similar to Example $4.14$. Then translate to a system of equations and solve it.

Step-by-Step Solution

Answer

The two required numbers are  $\frac{5}{2}$ and  $-\frac{5}{2}$.

Step 1. Identify the required numbers.

Lets take an example:

Assume $x$ and $y$ are the two numbers.

The sum of the two numbers is $0$.

So, $x+y=0 ...... \left(i\right)$

One number is five less than the other.

So, $y=x-5 ....... \left(ii\right)$

Therefore, the system of equations is  $$\begin{gathered} x+y=0 \\ y=x-5 \end{gathered}$$.

Step 2. Substitute the value of y in equation (i).

Substitute the value of $y$ in equation (i),

$$\begin{gathered} x+x-5=0 \\ 2x=5 \\ x=\frac{5}{2} \end{gathered}$$

Substituting $x=\frac{5}{2}$ in equation (ii),

$$\begin{gathered} y=\frac{5}{2}-5 \\ y=\frac{5-10}{2} \\ y=-\frac{5}{2} \end{gathered}$$

Step 3. Check the answer by substituting the values of x , y in the equations.

Substitute in equation (i),

$$\begin{gathered} x+y=0 \\ \frac{5}{2}+\left(-\frac{5}{2}\right)=0 \\ 0=0 \end{gathered}$$

This is true.

Again, substitute in equation (ii),

$$\begin{gathered} y=x-5 \\ -\frac{5}{2}=\frac{5}{2}-5 \\ -\frac{5}{2}=\frac{5-10}{5} \\ -\frac{5}{2}=-\frac{5}{2} \end{gathered}$$

This is also true.

The two required numbers are $\frac{5}{2}$ and $-\frac{5}{2}$.